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Problem of the Week #213 - Jun 28, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
Here is this week's POTW:

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Let $A$ be a commutative ring with unity. Prove that a polynomial $p(x) = a_0 + a_1 x + \cdots + a_n x^n$ over $A$ is a unit in $A[x]$ if and only if $a_0$ is an $A$-unit and $a_1,\ldots, a_n$ are nilpotent in $A$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
No one answered this week's problem. You can read my solution below.




Suppose $a_0$ is an $A$-unit and $a_1,\ldots, a_n$ are nilpotent in $A$. Then $a_1x, \ldots,..., a_n x^n$ are nilpotent in $A[x]$, so the sum $a_1 x + a_2 x^2 + ... + a_n x^n$ is nilpotent. That is, $p(x) - a_0$ is nilpotent in $A[x]$. Hence $p(x)$, being the sum of a unit and a nilpotent element, is a unit.

To prove the converse, let $p(x)$ is unit in $A[x]$. Let $q(x) = b_0 + b_1 x + ... + b_m x^m\in A[x]$ such that $p(x)q(x) = 1$. Suppose $q(x)$ is constant. Then $a_0$ is a unit as $a_0 b_0 = 1$; furthermore, $a_1 b_0 = a_2 b_0 = ... = a_n b_0 = 0$. Thus $a_1 = ... = a_n = 0$ and $a_1,..., a_n$ are nilpotent. If, on the other hand, $q(x)$ is nonconstant, then m ≥ 1 and the condition f(x) g(x) = 1 implies $a_0 b_0 = 1$ (making $a_0$ a unit) and $Pv = 0$, $P$ is an $(m + 1)\times (m + 1)$ matrix lower-triangular matrix with $a_n$'s along the diagonal, and where $v = \begin{pmatrix}b_m & b_{m-1} & \cdots & b_0\end{pmatrix}^T$. So $\operatorname{det}(P)v = 0$, or $a_n^m v = 0$. Since v is nonzero, an^m = 0. This proves $a_n$ is nilpotent. It follows that $q(x) - a_n x^n$, being the difference of a unit and a nilpotent, is a unit. Continuing this process we deduce nilpotency of $a_1,\ldots, a_{n-1}$.
 
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